Polar, cylindrical and spherical coordinates are special cases of one idea: describe the region with new variables in which it is simple, and correct for how the change stretches area. The correction is a determinant, the Jacobian.
Key idea
If $x = x(u, v)$, $y = y(u, v)$ maps a region $S$ of the $uv$-plane one-to-one onto $R$, then
$$\iint_R f(x, y)\,dx\,dy = \iint_S f(x(u, v), y(u, v))\left|\frac{\partial(x, y)}{\partial(u, v)}\right| du\,dv, \qquad \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix}.$$
Why it works
A small rectangle $\Delta u$ by $\Delta v$ in the $uv$-plane maps to a small curved patch. Near a point, the map is nearly linear: moving $\Delta u$ moves you by $\mathbf{r}_u\Delta u = \langle x_u, y_u \rangle\Delta u$, and moving $\Delta v$ by $\mathbf{r}_v\Delta v$. So the patch is nearly the parallelogram on those two vectors, with area $|x_uy_v - x_vy_u|\,\Delta u\,\Delta v$ (a 2-D cross product, which is the determinant). That number is the local stretch factor. Area cannot be negative, so take the absolute value. For polar coordinates, $\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r$: the familiar $dA = r\,dr\,d\theta$.
Worked examples
Worked example: A parallelogram
Evaluate $\iint_R (x + y)\,dA$ where $R$ is ${}0 \le x + y \le 2$, ${}0 \le x - y \le 4$.
- Let $u = x + y$ and $v = x - y$: $R$ becomes the rectangle ${}0 \le u \le 2$, ${}0 \le v \le 4$. Solving, $x = \frac{u + v}{2}$, $y = \frac{u - v}{2}$.
- $\frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{2}$, so $dA = \frac{1}{2}\,du\,dv$.
- $\int_0^4\int_0^2 u \cdot \frac{1}{2}\,du\,dv = \frac{1}{2} \cdot 2 \cdot 4 = 4$.
Answer: ${}4$
Worked example: Area of an ellipse
Find the area inside $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
- Let $x = au$, $y = bv$: the ellipse becomes the unit disk $u^2 + v^2 \le 1$.
- $\frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} a & 0 \\ 0 & b \end{vmatrix} = ab$.
- Area $= \iint_{\text{disk}} ab\,du\,dv = ab \cdot \pi = \pi ab$.
Answer: $\pi ab$
Worked example: Using the inverse Jacobian
If you only know $u = x + y$, $v = x - 2y$, how do you get $\frac{\partial(x, y)}{\partial(u, v)}$?
- Compute the easy one: $\frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = -3$.
- The two maps undo each other, so their stretch factors multiply to ${}1$: $\frac{\partial(x, y)}{\partial(u, v)} = -\frac{1}{3}$.
- So $dA = \frac{1}{3}\,du\,dv$.
Answer: $-\frac{1}{3}$, so $dA = \frac{1}{3}\,du\,dv$
Common mistake
Dropping the absolute value, or using the reciprocal. A negative Jacobian only means the map flips orientation; area is still positive, so use $\left|\frac{\partial(x, y)}{\partial(u, v)}\right|$. And the factor that goes with $du\,dv$ is the Jacobian of $x, y$ with respect to $u, v$. If you computed $\frac{\partial(u, v)}{\partial(x, y)}$ instead, take its reciprocal. Check with a simple case: the map $x = 2u$, $y = 2v$ must multiply areas by ${}4$.